The value of an item depreciates $x\%$ per year. The present value is $₹ 109350$, and after 3 years it will be $₹ 79716.15$. Using the depreciation formula: $A = P\left(1 - \frac{x}{100}\right)^n$ where $A$ is the final value, $P$ is the initial value, $x$ is the rate of depreciation, and $n$ is the time in years. <hr class="my-[18px]" node="[object Object]"> $79716.15 = 109350 \left(1 - \frac{x}{100}\right)^3$ $\left(1 - \frac{x}{100}\right)^3 = \frac{79716.15}{109350}$ $\left(1 - \frac{x}{100}\right)^3 = \frac{7971615}{10935000}$ $\left(1 - \frac{x}{100}\right)^3 = \frac{729}{1000}$ <hr class="my-[18px]" node="[object Object]"> Taking the cube root: $1 - \frac{x}{100} = \sqrt[3]{\frac{729}{1000}}$ $1 - \frac{x}{100} = \frac{9}{10}$ $1 - \frac{x}{100} = 0.9$ <hr class="my-[18px]" node="[object Object]"> Let the value of the item 2 years ago be $V$. The present value is the result of depreciating $V$ for 2 years: $109350 = V (0.9)^2$ $109350 = V (0.81)$ $V = \frac{109350}{0.81}$ $V = \frac{10935000}{81}$ $V = 135000$ The value of the item 2 years ago was $₹ 135000$.

Let the equation of the curve be $y = f(x)$. $y = x^2 - 4\cos x$ The slope of the tangent to the curve at any point $x$ is given by the first derivative, $\frac{dy}{dx}$. Let the slope be denoted by $m$. $m = \frac{dy}{dx}$ $m = \frac{d}{dx}(x^2 - 4\cos x)$ $m = 2x - 4(-\sin x)$ $m = 2x + 4\sin x$ To maximize this function $m$ on the interval $[0, \pi]$, differentiate $m$ with respect to $x$ and equate it to zero. <hr class="my-[18px]" node="[object Object]"> $\frac{dm}{dx} = \frac{d}{dx}(2x + 4\sin x)$ $\frac{dm}{dx} = 2 + 4\cos x$ Setting $\frac{dm}{dx} = 0$ for critical points: $2 + 4\cos x = 0$ $4\cos x = -2$ $\cos x = -\frac{1}{2}$ <hr class="my-[18px]" node="[object Object]"> Finding $x$ in the interval $[0, \pi]$ such that $\cos x = -\frac{1}{2}$: Since $\cos x$ is negative, $x$ must be in the second quadrant. Given that $\cos(\frac{\pi}{3}) = \frac{1}{2}$: $x = \pi - \frac{\pi}{3}$ $x = \frac{2\pi}{3}$ <hr class="my-[18px]" node="[object Object]"> To confirm that this value of $x$ gives a maximum slope, find the second derivative of $m$: $\frac{d^2m}{dx^2} = \frac{d}{dx}(2 + 4\cos x)$ $\frac{d^2m}{dx^2} = -4\sin x$ Evaluating at $x = \frac{2\pi}{3}$: $\frac{d^2m}{dx^2}\bigg|_{x=\frac{2\pi}{3}} = -4\sin\left(\frac{2\pi}{3}\right)$ $\frac{d^2m}{dx^2}\bigg|_{x=\frac{2\pi}{3}} = -4\left(\frac{\sqrt{3}}{2}\right)$ $\frac{d^2m}{dx^2}\bigg|_{x=\frac{2\pi}{3}} = -2\sqrt{3}$ Since the second derivative is negative ($&lt; 0$), the slope $m$ is at a local maximum at $x = \frac{2\pi}{3}$. <hr class="my-[18px]" node="[object Object]"> Checking the endpoints of the interval $[0, \pi]$ to ensure it is the global maximum: At $x = 0$: $m = 2(0) + 4\sin(0)$ $m = 0$ At $x = \pi$: $m = 2\pi + 4\sin(\pi)$ $m = 2\pi \approx 6.28$ At $x = \frac{2\pi}{3}$: $m = 2\left(\frac{2\pi}{3}\right) + 4\left(\frac{\sqrt{3}}{2}\right)$ $m = \frac{4\pi}{3} + 2\sqrt{3} \approx 7.65$ The value at $x = \frac{2\pi}{3}$ is the highest. Therefore, the slope of the tangent to the curve is maximum when $x = \frac{2\pi}{3}$.

Simplify the expression inside the inverse tangent using half-angle trigonometric identities: $1 - \cos x = 2\sin^2\left(\frac{x}{2}\right)$ $1 + \cos x = 2\cos^2\left(\frac{x}{2}\right)$ Substitute these into the function: $f(x) = \tan^{-1}\left(\sqrt{\frac{2\sin^2(x/2)}{2\cos^2(x/2)}}\right)$ $f(x) = \tan^{-1}\left(\sqrt{\tan^2\left(\frac{x}{2}\right)}\right)$ $f(x) = \tan^{-1}\left(\left|\tan\left(\frac{x}{2}\right)\right|\right)$ Since $x \in (0, \pi)$, the half-angle $\frac{x}{2} \in (0, \frac{\pi}{2})$. In this interval, $\tan(\frac{x}{2})$ is positive, so $|\tan(\frac{x}{2})| = \tan(\frac{x}{2})$. $f(x) = \tan^{-1}\left(\tan\left(\frac{x}{2}\right)\right)$ $f(x) = \frac{x}{2}$ <hr class="my-[18px]" node="[object Object]"> The normal is at $x = \frac{\pi}{3}$. The $y$-coordinate at this point: $y = f\left(\frac{\pi}{3}\right)$ $y = \frac{\pi/3}{2}$ $y = \frac{\pi}{6}$ The point of tangency is $\left(\frac{\pi}{3}, \frac{\pi}{6}\right)$. <hr class="my-[18px]" node="[object Object]"> The derivative of the simplified function $f(x) = \frac{x}{2}$: $f'(x) = \frac{1}{2}$ This is the slope of the tangent. The slope of the normal is the negative reciprocal of the tangent slope: $m_n = -\frac{1}{1/2}$ $m_n = -2$ <hr class="my-[18px]" node="[object Object]"> Using the point-slope form with point $\left(\frac{\pi}{3}, \frac{\pi}{6}\right)$ and slope $-2$: $y - \frac{\pi}{6} = -2\left(x - \frac{\pi}{3}\right)$ $y - \frac{\pi}{6} = -2x + \frac{2\pi}{3}$ $2x + y = \frac{2\pi}{3} + \frac{\pi}{6}$ $2x + y = \frac{4\pi + \pi}{6}$ $2x + y = \frac{5\pi}{6}$ <hr class="my-[18px]" node="[object Object]"> Check which of the given points satisfies the equation $2x + y = \frac{5\pi}{6}$: Option 1: $\left(\frac{\pi}{3}, \frac{\pi}{2}\right)$ $2\left(\frac{\pi}{3}\right) + \frac{\pi}{2} = \frac{7\pi}{6} \neq \frac{5\pi}{6}$ Option 2: $\left(\frac{\pi}{3}, 0\right)$ $2\left(\frac{\pi}{3}\right) + 0 = \frac{4\pi}{6} \neq \frac{5\pi}{6}$ Option 3: $(0, 0)$ $0 + 0 = 0 \neq \frac{5\pi}{6}$ Option 4: $\left(\frac{\pi}{3}, \frac{\pi}{6}\right)$ $2\left(\frac{\pi}{3}\right) + \frac{\pi}{6} = \frac{4\pi}{6} + \frac{\pi}{6} = \frac{5\pi}{6}$ ✓ Therefore, the normal passes through the point $\left(\frac{\pi}{3}, \frac{\pi}{6}\right)$.

The standard property of set theory states that for any sets $A, B, C,$ and $D$: $(A \times B) \cap (C \times D) = (A \cap C) \times (B \cap D)$ In this case, replacing $C$ with $B$ and $D$ with $A$: $(A \times B) \cap (B \times A) = (A \cap B) \times (B \cap A)$ <hr class="my-[18px]" node="[object Object]"> The intersection of sets is commutative, meaning $A \cap B = B \cap A$. Therefore: $(A \times B) \cap (B \times A) = (A \cap B) \times (A \cap B)$ <hr class="my-[18px]" node="[object Object]"> The cardinality of a Cartesian product $X \times Y$ is the product of the cardinalities of the individual sets: $n(X \times Y) = n(X) \times n(Y)$ Applying this to the simplified expression: $n((A \times B) \cap (B \times A)) = n((A \cap B) \times (A \cap B))$ $n((A \times B) \cap (B \times A)) = n(A \cap B) \times n(A \cap B)$ $n((A \times B) \cap (B \times A)) = [n(A \cap B)]^2$ <hr class="my-[18px]" node="[object Object]"> Given that $n(A \cap B) = 10$: $n((A \times B) \cap (B \times A)) = 10 \times 10$ $n((A \times B) \cap (B \times A)) = 100$ Therefore, the cardinality of $(A \times B) \cap (B \times A)$ is $100$.

The function is $g(x) = (\sqrt{3}\cos x + \sin x) - 1$. The expression $a\cos x + b\sin x$ ranges from $-\sqrt{a^2+b^2}$ to $\sqrt{a^2+b^2}$. Here, $a = \sqrt{3}$ and $b = 1$. Maximum value $= \sqrt{(\sqrt{3})^2 + 1^2}$ $= \sqrt{3+1}$ $= 2$ Minimum value $= -2$ So, the range of $(\sqrt{3}\cos x + \sin x)$ is $[-2, 2]$. Therefore, the range of $g(x)$ is $[-2 - 1, 2 - 1]$, which simplifies to $[-3, 1]$. <hr class="my-[18px]" node="[object Object]"> Let $u = g(x)$, so $u \in [-3, 1]$. The function becomes $f(u) = \left|\frac{u}{2}\right|$. Analyzing the values of $f(u)$ as $u$ varies from $-3$ to $1$: If $u = -3$, $f(-3) = |-1.5| = 1.5$ (Maximum value) If $u = 0$, $f(0) = 0$ (Minimum value) If $u = 1$, $f(1) = 0.5$ The range of the composite function $(f \circ g)(x)$ is continuous from the minimum to the maximum, which is $[0, 1.5]$. <hr class="my-[18px]" node="[object Object]"> For any value $y$ in the interval $[0, 1.5]$, the greatest integer function gives: For $y \in [0, 1)$, $[y] = 0$ For $y \in [1, 1.5]$, $[y] = 1$ Thus, the possible output values are only the integers $0$ and $1$. Therefore, the range is $\{0, 1\}$.

Perimeter of a regular hexagon = Perimeter of a regular octagon. Value to use: $\tan 67.5^\circ = (\sqrt{2} + 1)$. Let the side length of the regular hexagon be $a$ and the side length of the regular octagon be $b$. Since their perimeters are equal: $6a = 8b$ $a = \frac{8b}{6}$ $a = \frac{4b}{3}$ <hr class="my-[18px]" node="[object Object]"> The area of a regular hexagon with side $a$ is given by: $A_H = \frac{3\sqrt{3}}{2} a^2$ Substituting $a = \frac{4b}{3}$: $A_H = \frac{3\sqrt{3}}{2} \left( \frac{4b}{3} \right)^2$ $A_H = \frac{3\sqrt{3}}{2} \times \frac{16b^2}{9}$ $A_H = \frac{8\sqrt{3}}{3} b^2$ <hr class="my-[18px]" node="[object Object]"> A regular octagon can be divided into 8 congruent isosceles triangles with the center of the octagon as the common vertex. Base of each triangle = side of octagon = $b$ Central angle = $\frac{360^\circ}{8} = 45^\circ$ Base angles of the triangle = $\frac{180^\circ - 45^\circ}{2} = 67.5^\circ$ Let $h$ be the height (apothem) of one such triangle. Using trigonometry in the right-angled half of the triangle: $\tan 67.5^\circ = \frac{h}{(b/2)}$ $h = \frac{b}{2} \tan 67.5^\circ$ Area of one triangle: $= \frac{1}{2} \times b \times \frac{b}{2} \tan 67.5^\circ$ $= \frac{b^2}{4} \tan 67.5^\circ$ Total Area of Octagon: $A_O = 8 \times \frac{b^2}{4} \tan 67.5^\circ$ $A_O = 2b^2 \tan 67.5^\circ$ Using $\tan 67.5^\circ = (\sqrt{2} + 1)$: $A_O = 2b^2 (\sqrt{2} + 1)$ <hr class="my-[18px]" node="[object Object]"> The ratio $\frac{A_O}{A_H}$: $\text{Ratio} = \frac{2b^2(\sqrt{2} + 1)}{\frac{8\sqrt{3}}{3}b^2}$ $\text{Ratio} = \frac{2(\sqrt{2} + 1) \times 3}{8\sqrt{3}}$ $\text{Ratio} = \frac{6(\sqrt{2} + 1)}{8\sqrt{3}}$ $\text{Ratio} = \frac{3(\sqrt{2} + 1)}{4\sqrt{3}}$ Rationalizing the denominator: $\text{Ratio} = \frac{3\sqrt{3}(\sqrt{2} + 1)}{4 \times 3}$ $\text{Ratio} = \frac{\sqrt{3}(\sqrt{2} + 1)}{4}$ $\text{Ratio} = \frac{\sqrt{6} + \sqrt{3}}{4}$

The relevant data from the table: <table class="question-table"><thead><tr><th style="text-align: left;">School</th><th style="text-align: left;">Class</th><th style="text-align: left;">Total Students</th><th style="text-align: left;">% of Boys</th><th style="text-align: left;">% of Girls</th></tr></thead><tbody><tr><td node="[object Object]" style="text-align: left;">S3</td><td node="[object Object]" style="text-align: left;">XI</td><td node="[object Object]" style="text-align: left;">100</td><td node="[object Object]" style="text-align: left;">65%</td><td node="[object Object]" style="text-align: left;">$100 - 65 = 35\%$</td></tr><tr><td node="[object Object]" style="text-align: left;">S2</td><td node="[object Object]" style="text-align: left;">IX</td><td node="[object Object]" style="text-align: left;">150</td><td node="[object Object]" style="text-align: left;">30%</td><td node="[object Object]" style="text-align: left;">—</td></tr></tbody></table> <hr class="my-[18px]" node="[object Object]"> The number of girls in School S3 (Class XI): $\text{Girls}$ $= \text{Total Students} \times (\% \text{ Girls})$ $= 100 \times 35\%$ $= 100 \times 0.35$ $= 35$ <hr class="my-[18px]" node="[object Object]"> The number of boys in School S2 (Class IX): $\text{Boys}$ $= \text{Total Students} \times (\% \text{ Boys})$ $= 150 \times 30\%$ $= 150 \times 0.30$ $= 45$ <hr class="my-[18px]" node="[object Object]"> The percentage that the number of girls in S3 (Class XI) represents of the number of boys in S2 (Class IX): $\text{Percentage}$ $= \left( \frac{\text{Girls (S3, XI)}}{\text{Boys (S2, IX)}} \right) \times 100$ $= \left( \frac{35}{45} \right) \times 100$ $= \frac{7}{9} \times 100$ $\approx 77.78\%$ Therefore, the answer is 77.78 (Option 3).

Calculate the number of girls in Class XII for all schools to satisfy the condition: $\text{Girls}(A) &gt; \text{Girls}(B) + \text{Girls}(C)$ <hr class="my-[18px]" node="[object Object]"> Using the formula: $\text{Girls} = \text{Total} \times (100\% - \% \text{Boys})$ <table class="question-table"><thead><tr><th style="text-align: left;">School</th><th style="text-align: left;">Total Students (XII)</th><th style="text-align: left;">% Boys</th><th style="text-align: left;">% Girls</th><th style="text-align: left;">Calculation</th><th style="text-align: left;">No. of Girls</th></tr></thead><tbody><tr><td node="[object Object]" style="text-align: left;">S1</td><td node="[object Object]" style="text-align: left;">125</td><td node="[object Object]" style="text-align: left;">80%</td><td node="[object Object]" style="text-align: left;">20%</td><td node="[object Object]" style="text-align: left;">$125 \times 0.20$</td><td node="[object Object]" style="text-align: left;">25</td></tr><tr><td node="[object Object]" style="text-align: left;">S2</td><td node="[object Object]" style="text-align: left;">100</td><td node="[object Object]" style="text-align: left;">45%</td><td node="[object Object]" style="text-align: left;">55%</td><td node="[object Object]" style="text-align: left;">$100 \times 0.55$</td><td node="[object Object]" style="text-align: left;">55</td></tr><tr><td node="[object Object]" style="text-align: left;">S3</td><td node="[object Object]" style="text-align: left;">120</td><td node="[object Object]" style="text-align: left;">60%</td><td node="[object Object]" style="text-align: left;">40%</td><td node="[object Object]" style="text-align: left;">$120 \times 0.40$</td><td node="[object Object]" style="text-align: left;">48</td></tr><tr><td node="[object Object]" style="text-align: left;">S4</td><td node="[object Object]" style="text-align: left;">150</td><td node="[object Object]" style="text-align: left;">40%</td><td node="[object Object]" style="text-align: left;">60%</td><td node="[object Object]" style="text-align: left;">$150 \times 0.60$</td><td node="[object Object]" style="text-align: left;">90</td></tr></tbody></table> <hr class="my-[18px]" node="[object Object]"> Checking each option to find where the first school has more girls than the sum of the other two: Option 1: S4, S1, S2 $\text{Girls}(S4) &gt; \text{Girls}(S1) + \text{Girls}(S2)$ $90 &gt; 25 + 55$ $90 &gt; 80$ This is TRUE. <hr class="my-[18px]" node="[object Object]"> Option 2: S2, S1, S4 $55 &gt; 25 + 90$ $55 &gt; 115$ This is FALSE. <hr class="my-[18px]" node="[object Object]"> Option 3: S2, S1, S3 $55 &gt; 25 + 48$ $55 &gt; 73$ This is FALSE. <hr class="my-[18px]" node="[object Object]"> Option 4: S4, S2, S3 $90 &gt; 55 + 48$ $90 &gt; 103$ This is FALSE. <hr class="my-[18px]" node="[object Object]"> Therefore, the answer is Option 1: S4, S1, S2.

The number of boys for each school is calculated by multiplying the Total Students by the Percentage of Boys. <table class="question-table"><thead><tr><th style="text-align: left;">School</th><th style="text-align: left;">Class IX Boys</th><th style="text-align: left;">Class X Boys</th><th style="text-align: left;">Class XI Boys</th><th style="text-align: left;">Class XII Boys</th></tr></thead><tbody><tr><td node="[object Object]" style="text-align: left;">S1</td><td node="[object Object]" style="text-align: left;">$120 \times 40\% = 48$</td><td node="[object Object]" style="text-align: left;">$90 \times 30\% = 27$</td><td node="[object Object]" style="text-align: left;">$150 \times 60\% = 90$</td><td node="[object Object]" style="text-align: left;">$125 \times 80\% = 100$</td></tr><tr><td node="[object Object]" style="text-align: left;">S2</td><td node="[object Object]" style="text-align: left;">$150 \times 30\% = 45$</td><td node="[object Object]" style="text-align: left;">$100 \times 55\% = 55$</td><td node="[object Object]" style="text-align: left;">$120 \times 50\% = 60$</td><td node="[object Object]" style="text-align: left;">$100 \times 45\% = 45$</td></tr><tr><td node="[object Object]" style="text-align: left;">S3</td><td node="[object Object]" style="text-align: left;">$140 \times 20\% = 28$</td><td node="[object Object]" style="text-align: left;">$120 \times 40\% = 48$</td><td node="[object Object]" style="text-align: left;">$100 \times 65\% = 65$</td><td node="[object Object]" style="text-align: left;">$120 \times 60\% = 72$</td></tr><tr><td node="[object Object]" style="text-align: left;">S4</td><td node="[object Object]" style="text-align: left;">$100 \times 35\% = 35$</td><td node="[object Object]" style="text-align: left;">$100 \times 45\% = 45$</td><td node="[object Object]" style="text-align: left;">$80 \times 80\% = 64$</td><td node="[object Object]" style="text-align: left;">$150 \times 40\% = 60$</td></tr><tr><td node="[object Object]" style="text-align: left;">TOTAL</td><td node="[object Object]" style="text-align: left;">156</td><td node="[object Object]" style="text-align: left;">175</td><td node="[object Object]" style="text-align: left;">279</td><td node="[object Object]" style="text-align: left;">277</td></tr></tbody></table> <hr class="my-[18px]" node="[object Object]"> The total number of boys in each class across all schools: Class IX: $156$ Class X: $175$ Class XI: $279$ Class XII: $277$ <hr class="my-[18px]" node="[object Object]"> Arranging the classes from lowest to highest: $156 &lt; 175 &lt; 277 &lt; 279$ The correct arrangement is IX, X, XII, XI.

Since the table gives the percentage of boys, the percentage of girls is calculated as: $\% \text{Girls} = 100\% - \% \text{Boys}$ The number of girls for each class is calculated using: $\text{Number of Girls} = \text{Total Students} \times \% \text{Girls}$ <hr class="my-[18px]" node="[object Object]"> The table below breaks down the calculation for each class and sums the total girls for every school. <table class="question-table"><thead><tr><th style="text-align: left;">School</th><th style="text-align: left;">Class IX</th><th style="text-align: left;">Class X</th><th style="text-align: left;">Class XI</th><th style="text-align: left;">Class XII</th><th style="text-align: left;">Total Girls</th></tr></thead><tbody><tr><td node="[object Object]" style="text-align: left;">S1</td><td node="[object Object]" style="text-align: left;">$120 \times 60\% = 72$</td><td node="[object Object]" style="text-align: left;">$90 \times 70\% = 63$</td><td node="[object Object]" style="text-align: left;">$150 \times 40\% = 60$</td><td node="[object Object]" style="text-align: left;">$125 \times 20\% = 25$</td><td node="[object Object]" style="text-align: left;">220</td></tr><tr><td node="[object Object]" style="text-align: left;">S2</td><td node="[object Object]" style="text-align: left;">$150 \times 70\% = 105$</td><td node="[object Object]" style="text-align: left;">$100 \times 45\% = 45$</td><td node="[object Object]" style="text-align: left;">$120 \times 50\% = 60$</td><td node="[object Object]" style="text-align: left;">$100 \times 55\% = 55$</td><td node="[object Object]" style="text-align: left;">265</td></tr><tr><td node="[object Object]" style="text-align: left;">S3</td><td node="[object Object]" style="text-align: left;">$140 \times 80\% = 112$</td><td node="[object Object]" style="text-align: left;">$120 \times 60\% = 72$</td><td node="[object Object]" style="text-align: left;">$100 \times 35\% = 35$</td><td node="[object Object]" style="text-align: left;">$120 \times 40\% = 48$</td><td node="[object Object]" style="text-align: left;">267</td></tr><tr><td node="[object Object]" style="text-align: left;">S4</td><td node="[object Object]" style="text-align: left;">$100 \times 65\% = 65$</td><td node="[object Object]" style="text-align: left;">$100 \times 55\% = 55$</td><td node="[object Object]" style="text-align: left;">$80 \times 20\% = 16$</td><td node="[object Object]" style="text-align: left;">$150 \times 60\% = 90$</td><td node="[object Object]" style="text-align: left;">226</td></tr></tbody></table> <hr class="my-[18px]" node="[object Object]"> Arranging the schools from the highest number of girls to the lowest: S3: 267 S2: 265 S4: 226 S1: 220 Therefore, the descending order is: $S3, S2, S4, S1$

iimb-ugat 2025 Complete Paper Solution | QA

Question 1.

Question 2.

Question 3.

Question 4.

Question 5.

Question 6.

Instructions

Question 7.

Question 8.

Question 9.

Question 10.

iimb-ugat 2025 Complete Paper Solution | QA

Question 1.

<div><div class="latex-wrapper">The value of an item depreciates $x\%$ per year. The present value of the item is $₹ 109350$. Its value after 3 years will be $₹ 79716.15$. Two years ago from now, its value, in rupees, was<br /></div></div>

Question 2.

<div><div class="latex-wrapper">The slope of the tangent to the curve $y = x^2 - 4\cos x$, $x \in [0, \pi]$ is maximum when $x$ is equal to<br /></div></div>

Question 3.

<div><div class="latex-wrapper">Let $f(x) = \tan^{-1}\left(\sqrt{\frac{1-\cos x}{1+\cos x}}\right)$; $x \in (0,\pi)$. A normal to $y = f(x)$ at $x = \frac{\pi}{3}$ passes through the point:<br /></div></div>

Question 4.

<div><div class="latex-wrapper">Let $A \times B$ denote the Cartesian product of sets $A$ and $B$. If the cardinality of $A \cap B$ is $10$, then the cardinality of $(A \times B) \cap (B \times A)$ is:<br /></div></div>

Question 5.

<div><div class="latex-wrapper">Let $f(x) = \left|\frac{x}{2}\right|$ and $g(x) = \sqrt{3}\cos x + \sin x - 1$, for all real numbers $x$. If $[ \ \ ]$ denotes the greatest integer function, then the range of the function $[f \circ g]$ is:<br /></div></div>

Question 6.

<div><div class="latex-wrapper">The perimeters of a regular hexagon and a regular octagon are equal. Find the ratio of the area of the octagon to that of the hexagon. [Use $\tan 67.5° = (\sqrt{2} + 1)$]<br /></div></div>

Instructions

Question 7.

<div><div class="latex-wrapper">The number of girls in XI in the school S3 is approximately ______% of the number of boys in IX in the school S2.<br /></div></div>

Question 8.

<div><div class="latex-wrapper">In class XII, the number of girls in the school ______ is more than the combined number of girls in the school ______ and ______.<br /></div></div>

Question 9.

<div><div class="latex-wrapper">For all schools, the correct arrangement of classes in ascending order of number of boys is ______.<br /></div></div>

Question 10.

<div><div class="latex-wrapper">On arranging schools in descending order of total number of girls in all classes, the correct order is ______.<br /></div></div>